To determine the nature of the solutions for the quadratic polynomial \(-2x^2 + 4x - 9 = 0\), we'll use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
For our specific polynomial, the coefficients are:
- \(a = -2\)
- \(b = 4\)
- \(c = -9\)
Now, we'll substitute these values into the quadratic formula.
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Calculate the discriminant \(b^2 - 4ac\): \[ b^2 = 4^2 = 16 \] \[ 4ac = 4 \cdot (-2) \cdot (-9) = 72 \] \[ b^2 - 4ac = 16 - 72 = -56 \]
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Since the discriminant is negative (\(-56\)), we will have two non-real solutions.
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Now we can calculate the solutions using the quadratic formula: \[ x = \frac{-4 \pm \sqrt{-56}}{2(-2)} \] Simplifying further: \[ x = \frac{-4 \pm \sqrt{56}i}{-4} = \frac{4 \mp \sqrt{56}i}{4} \] \[ x = 1 \mp \frac{\sqrt{56}}{4} i \]
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Therefore, the solutions are: \[ x = 1 + \frac{\sqrt{56}}{4} i \quad \text{and} \quad x = 1 - \frac{\sqrt{56}}{4} i \]
As the solutions are complex (non-real), the correct statement about the solution set is:
There are two non-real solutions.